# Intuition behind showing that $\{x ∈ \mathbb{R}$ $: f(x) > 0\}$ is countable for $\sum f(x)< \infty$

Problem

In a book on Measure Theory, I encountered the following problem:

Let $$f : \mathbb{R} → [0,+∞]$$ and suppose that $$\sum _{x \in \mathbb{R}} f(x) < ∞$$.

Show that the set $$\{x ∈ \mathbb{R} : f(x) > 0\}$$ is countable.

The solution is provided, although I find the general direction it takes to be quite unintuitive and unclear in parts too. The beginning of the solution is written below alongside my question below it.

Let $$M := \sum _{x∈R} f(x) < ∞$$. We claim the set {$$x ∈ R : f(x) > \frac{M}{k}$$} has at most k elements, hence it is finite.

What is the intuition behind this? When attempting this type of problem, is there a way in which I could have been motivated to make the above conjecture?

The proof itself makes sense when I look through the rest of it, but I find it unlikely that I would have stumbled upon the answer myself from this. Is this just a trick or there is something to learn from this example that I can take forward when considering similar problems of this nature?

Terminology

To clarify for those unsure about the definition used for the sum over the real numbers, we use the following definition:

$$\sum _{x \in \mathbb{R}} f(x) = \sup _{\{\text{finite} \space A \subseteq \mathbb{R}\}} \sum_{x \in A}f(x)$$

(as mentioned in the comments by PrincessEev).

• How do you define the sum over an uncountable index set? Commented Jan 4, 2023 at 7:26
• The only definition I've seen is $\sup_F \sum_{x \in F} f(x)$ for $F$ a finite subset of the original indexing set. Commented Jan 4, 2023 at 7:33
• I don't know if I'd have found this myself without prompting, but I think the idea probably comes from considering the idea that if the lower bound of $0$ were tighter, each set $S_\epsilon = \{x \in \mathbb R : f(x) > \epsilon > 0\}$ must be finite, since otherwise $\sum_{\mathbb R} f(x) \geq \sum_{S_\epsilon} f(x) \geq \epsilon \cdot \sum_{S_\epsilon} 1 = \epsilon \cdot \infty.$ Then from there it would be sensible to try to find a countable sequence of $\epsilon$'s which go to zero, and take the union. Commented Jan 4, 2023 at 7:35
• The (unusual) notation $\sum_{x\in\mathbb R}f(x)$ could be interpreted also as $\int fd\mu$ where $\mu$ denotes the counting measure on $\mathcal P(\mathbb R)$ which is prescribed by $A\mapsto|A|$. Commented Jan 4, 2023 at 9:12
• I believe that this particular proof would be almost completely unaltered if we were to use this definition. I haven't worked through it, but I don't see why this would result in a major difference in approaches @drhab Commented Jan 4, 2023 at 9:16

From $$\sum_{x\in\mathbb R}f(x)<\infty$$ it follows directly that for every positive integer $$n$$ the set: $$A_n:=\left\{x\in\mathbb R\mid f(x)\geq\frac1n\right\}$$ is finite.

Then based on: $$\{x\in\mathbb R\mid f(x)>0\}=\bigcup_{n=1}^{\infty}A_n$$we find directly that the set $$\{x\in\mathbb R\mid f(x)>0\}$$ is a union of finite sets hence is countable.

• I don't see how this is any different to my answer (in fact it clearly is), albeit, I prove the contrapositive. But it's the exact same idea... Commented Jan 4, 2023 at 10:37
• @AdamRubinson It is not essentially different but more concise I would say. This also because it avoids the use of a contradiction. Commented Jan 4, 2023 at 10:48
• Yes that's fair. Commented Jan 4, 2023 at 10:48

The motivation is essentially that, since we know the range of the function and hence the summation is positive, we can break up the range's values. Several different ways could probably apply, but ultimately the idea here is that

• we want a countable collection of "breaking points".
• we want those "breaking points" to have upper bound $$M$$ and lower bound $$0$$; these have to be the bounds, since if $$f(x)>M$$ for any single $$x$$, then the sum exceeds $$M$$, and we know that $$f(x) \ge 0$$ for all $$x$$ by definition.
• we want those "breaking points" to be arranged in such a way that the union of the sets they generate is $$\{x \mid f(x) > 0\}$$.

The "breaking points" in this scenario are the $$M/k$$'s; notice that this is decreasing in $$k$$ to $$0$$, so all three conditions are met. In particular,

$$\{x \mid f(x) > 0\} = \bigcup_{k \in \mathbb{N}} \left\{ x \, \middle| \, f(x) > \frac M k \right\}$$

This construction in particular works nicely for us, because the conclusion we reach (that each set in the union has at most $$k$$ elements) is almost obvious: if more than $$k$$ elements were in the set, then the sum would exceed $$M$$ on just those members of the set! Then it's as simple as the realization that the union of finite sets is at-most-countable.

The ultimate trick here is just a matter of partitioning up the range of the function, a common theme in analysis and in particular measure theory. Perhaps even after seeing it, it wouldn't be your first approach, but it becomes more natural the more you see this type of technique applied.

Suppose $$\ X:= \{x \in \mathbb{R} : f(x) > 0\}\$$ is uncountable.$$\qquad (1)$$

Define $$\ X_0:=\{x \in \mathbb{R} : f(x) \geq 1\},\$$ and for each $$\ n\in\mathbb{N},\$$ define

$$\ X_n:= \{x \in \mathbb{R} : \frac{1}{n+1}\leq f(x) < \frac{1}{n}\}.\$$

Suppose further that $$\ \vert X_n\vert\$$ is countably infinite for each $$\ n\in\mathbb{N}\cup\{0\}.\qquad (2)$$

Then $$\ X = \displaystyle\bigcup_{n\in\mathbb{N}\cup\{0\}} X_n\$$ would be the countable union of countably many sets, which is countable.

However, since we started assuming $$(1),\ X\$$ is uncountable. We have contradicted assumption $$(2),$$ and so under assumption $$(1),\ \exists\ k\in\mathbb{N}\$$ such that $$\ \vert X_k\vert\$$ is uncountable.

The result follows because, under assumption $$(1),\ \displaystyle\sum_{x\in\mathbb{R}} f(x) \geq \sum_{j\in\mathbb{N}} \frac{1}{k+1} = \infty.$$

It would have sufficed to show that $$\ \vert X_j\vert\$$ is countably infinite for some $$\ j\in\mathbb{N},\$$ but I've done one better.

• But then we would not have $\ X = \displaystyle\bigcup_{n\in\mathbb{N}\cup\{0\}} X_n,\$ right? So how would we make a similar argument without $X_0$ ? Commented Jan 4, 2023 at 8:31
• Actually, my last previous comment with $\ X_n:= \{x \in \mathbb{R} : n-1 < f(x) \leq n\}.\$ is wrong, so I've deleted it. The proof wouldn't work then. The reason I defined $\ X_n\$ the way I did in my answer is that each $\ X_n\$ is sectioned off away from zero. Since one of these "sectioned away from zero" sets has infinitely many points of $X$, therefore the sum of members of $X$ is infinity. Commented Jan 4, 2023 at 8:47
• I'll start by addressing for your second query, as it's to do with my overall argument, which is more important. I may be wrong (as I'm not sure exactly what you mean), but I suspect you don't follow the overall structure of my argument. It is this: $(1)$ Suppose $X$ is uncountable. $(2)$ Then there exists $k$ such that $X_k$ is uncountable. $(3)$ Therefore, $\ \displaystyle\sum_{x\in\mathbb{R}} f(x)$ diverges. $\qquad$ Or to summarise further, I have proven that $(1)\implies (3),\$ which is the contrapositive of the thing we are trying to prove. Does this address your concern? Commented Jan 5, 2023 at 7:34
• Next, you say, "I don't see where the sum of $\frac{1}{k+1}$ bound came from". Well, each member of $X_k$ satisfies $f(x) \geq \frac{1}{k+1}$ by definition. Since $\vert X_k\vert$ is (uncountable, and therefore) infinite, it follows that $\ f(x) \geq \frac{1}{k+1}$ for infinitely many $x,$ and therefore $\ \displaystyle\sum_{x\in\mathbb{R}} f(x) \geq \sum_{x\in X_k} f(x) \geq \infty \times \frac{1}{k+1} = \infty.$ Commented Jan 5, 2023 at 7:34